91Ó°ÊÓ

To find the total number of arrangements of the letters in UNUSUAL, we need to apply the formula for permutations of a multiset because there are repeating letters. The word 'UNUSUAL' has 7 letters with U repeating 3 times, S repeating once, N repeating once, A repeating once and L repeating once. So, the total number of arrangements is given by the formula \( \frac{n!}{n1!n2!...nj!} \) where \( n \) is the total number of letters, and each \( ni \) represents the number of times a specific letter appears in the word. In this case, \( \frac{7!}{3!1!1!1!1!} = 5,040 \) arrangements.

To find the arrangements with all the U's together, we can treat the three U's as a single entity. This would imply we now have 5 entities: 'UUU', 'N', 'S', 'A', 'L'. Therefore we now have to find the arrangements for 5 entities and this can be done in \( 5! = 120 \) ways.

Calculating the arrangements with no consecutive U's is more challenging. One way to approach this is by subtracting from the total arrangements those that have at least two U's together. From Step 1, we know that total arrangements are 5,040. Now we have to subtract the arrangements that have at least two U's together. We have already calculated the arrangements that have all U's together (120). Add to that the arrangements which have any two U's together. These can be found by considering any two U's as a single entity, so we have 6 entities and this can be done in \( \frac{6!}{2!1!1!1!1!} = 720 \) ways. So, total arrangements with at least two U's together are \( 720 + 120 = 840 \). Therefore, the arrangements with no consecutive U's are \( 5,040 - 840 = 4,200 \)